MATH383: Solving Homogeneous Linear Systems of Differential Equations (hLSDEs)

Overview

• Fundamental set of solutions (Wronskian)

• Constant-coefficient hLSDEs, matrix eigenvalue problem

• Cases:

  • Real negative eigenvalues

  • Real positive eigenvalues

  • Purely imaginary eigenvalues

  • Complex eigenvalues

hLSDE fundamental set of solutions, Wronskian

• ${𝒚}^{\text{'}}=𝑨\left(t\right)𝒚$, $𝒚\left(t\right)$, $𝒚:ℝ\to {ℝ}^n$

• Set of solutions $𝒮=\{{𝒚}_1,\dots ,{𝒚}_n\}$, ${𝒚}_i^{\text{'}}=𝑨{𝒚}_i$, $i=1,\dots ,n$

• $𝒮$ is a fundamental set of solutions of $\{{𝒚}_1\left(t\right),\dots ,{𝒚}_n\left(t\right)\}$ are linearly independent (for all $t$)

• $𝒚=c_1{𝒚}_1+\cdots +c_n{𝒚}_n$, $c_1{𝒚}_1+\cdots +c_n{𝒚}_n=𝟎\Rightarrow c_1=\cdots =c_n=0.$

  ${\displaystyle 𝒚=\left(\begin{array}{lll}{𝒚}_1& \dots & {𝒚}_n\end{array}\right)𝒄=\left(\begin{array}{llll}{y}_{11}& y_{12}& \dots & y_{1n}\\ y_{21}& y_{22}& \dots & y_{2n}\\ ⋮& ⋮& \ddots & ⋮\\ y_{n1}& y_{n2}& \dots & y_{nn}\end{array}\right)\left(\begin{array}{l}{c}_1\\ c_2\\ ⋮\\ c_n\end{array}\right)=𝒀𝒄=𝒚}$

• $𝒀𝒄=𝒚$ always has a solution iff $ℛ\left(𝒀\right)={ℝ}^n$, or equivalently if $\det \left(𝒀\right)\ne 0$

• The Wronskian of $𝒮=\{{𝒚}_1,\dots ,{𝒚}_n\}$ is $W\left(t\right)=\det \left(𝒀\left(t\right)\right)$,$W:ℝ\to ℝ$.

Theorem. Abel's Formula. If $𝑨\left(t\right)$ is continuous on $(a,b)$, and $𝒮=\{{𝒚}_1,\dots ,{𝒚}_n\}$ is a set of solutions of ${𝒚}^{\text{'}}=𝑨𝒚$ on $(a,b)$, then the Wronskian of $𝒮$ is given by

hLSDE, Wronskian

• $y:ℝ\to ℝ,y^{\text{'}}=a\left(t\right)y\Rightarrow y\left(t\right)=\exp [{\int }_{t_0}^{t}a\left(s\right)ds]y\left(t_0\right)$

  ${\displaystyle y^{\text{'}}\left(t\right)={(\underset{t_0}{\overset{t}{\int }}a\left(s\right)ds)}^{\text{'}}\exp [\underset{t_0}{\overset{t}{\int }}a\left(s\right)ds]y\left(t_0\right)=a\left(t\right)y}$ ${\displaystyle \exp \left(a\right)=\frac{1}{0!}{a}^0+\frac{1}{1!}{a}^1+\frac{1}{2!}{a}^2+\frac{1}{3!}{a}^3+\cdots =1+a+\frac{1}{2}{a}^2+\frac{1}{6}{a}^3+\cdots }$

• $𝒚:ℝ\to {ℝ}^n$, ${𝒚}^{\text{'}}=𝑨\left(t\right)𝒚$, $𝒚\left(t\right)=\exp [{\int }_{t_0}^t𝑨\left(s\right)ds]𝒚\left(t_0\right)$,

  ${\displaystyle {𝒚}^{\text{'}}\left(t\right)={(\underset{t_0}{\overset{t}{\int }}𝑨\left(s\right)ds)}^{\text{'}}\exp [\underset{t_0}{\overset{t}{\int }}𝑨\left(s\right)ds]𝒚\left(t_0\right)=𝑨\left(t\right)𝒚}$ ${\displaystyle \exp \left(𝑨\right)=\frac{1}{0!}{𝑨}^0+\frac{1}{1!}{𝑨}^1+\frac{1}{2!}{𝑨}^2+\frac{1}{3!}{𝑨}^3+\cdots =𝑰+𝑨+\frac{1}{2}{𝑨}^2+\frac{1}{6}{𝑨}^3+\cdots }$

• trace of a matrix

  ${\displaystyle \mathrm{tr}𝑨\left(t\right)=a_{11}\left(t\right)+\cdots +a_{nn}\left(t\right)=\sum _{i=1}^n{a}_{ii}\left(t\right)}$

Wronskian example

Example 1. Check is $𝒮=\{{𝒚}_1,{𝒚}_2\}$ is a fundamental set of solutions for ${𝒚}^{\text{'}}=𝑨𝒚$

Solution 1. Solve $c_1{𝒚}_1+c_2{𝒚}_2=𝟎$

Solution 2. Directly compute the Wronskian

Solution 3. Apply Abel's formula

Solution of constant-coefficient hLSDEs

• Consider ${𝒚}^{\text{'}}=𝑨𝒚$. How to find a solution?

• Recall that for $y^{\text{'}}=ay$, we'd guess $y\left(t\right)=c{e}^{\lambda t}\Rightarrow \lambda =a$, $y\left(t\right)=c{e}^{\lambda t}$

• Try the same approach for systems, guess that the solution is of the form

  ${\displaystyle 𝒚=e^{\lambda t}𝒙\Rightarrow \lambda e^{\lambda t}𝒙=𝑨(e^{\lambda t}𝒙)\Rightarrow 𝑨𝒙=\lambda 𝒙}$

• For $𝑨\in {ℝ}^{n\times n}$ finding $\lambda ,𝒙$ that satisfy $𝑨𝒙=\lambda 𝒙$ is known as the eigenvalue problem

  • $\lambda$ are the eigenvalues of $𝑨$

  • $𝒙$ are the eigenvectors of $𝑨$ ($𝒙\ne 𝟎$, no trivial solutions)

• The eigenvalue relationship can be rewritten as

  ${\displaystyle 𝑨𝒙=\lambda 𝒙\Rightarrow 𝑨𝒙-\lambda 𝒙=𝟎,(𝑨-\lambda 𝑰)𝒙=𝟎\Rightarrow p_{𝑨}\left(\lambda \right)=\det (\lambda 𝑰-𝑨)=0}$

  where $p_{𝑨}\left(\lambda \right)$ is the characteristic polynomial of $𝑨$, and is of degree $n$, with $n$ roots.

• Denote the $n$ eigenvalues as ${\lambda }_1,\dots ,{\lambda }_n$, and the $n$ eigenvectors as ${𝒙}_1,\dots ,{𝒙}_n$

Characteristic polynomial

• $p_{𝑨}\left(\lambda \right)=\det (\lambda 𝑰-𝑨)$

  ${\displaystyle p_{𝑨}\left(\lambda \right)=\left|\left(\begin{array}{llll}\lambda -a_{11}& a_{12}& \dots & a_{1n}\\ a_{21}& \lambda -a_{22}& \dots & a_{2n}\\ ⋮& ⋮& \ddots & ⋮\\ a_{n1}& a_{n2}& \dots & \lambda -a_{nn}\end{array}\right)\right|}$

• $p_{𝑨}\left(\lambda \right)={\lambda }^n-\mathrm{tr}𝑨{\lambda }^{n-1}+\cdots +{(-1)}^n\det 𝑨={\lambda }^n+a_{n-1}{\lambda }^{n-1}+\cdots +a_0$

• $p_{𝑨}\left(\lambda \right)=(\lambda -{\lambda }_1)(\lambda -{\lambda }_2)\dots (\lambda -{\lambda }_{n-1})(\lambda -{\lambda }_n)={\lambda }^n+(-{\lambda }_1-{\lambda }_2-\cdots -{\lambda }_n){\lambda }^{n-1}$

• Vieta relations

  ${\displaystyle \begin{array}{l}{\lambda }_1+{\lambda }_2+\cdots +{\lambda }_n=-a_{n-1}=\mathrm{tr}𝑨\\ {\lambda }_1{\lambda }_2+{\lambda }_1{\lambda }_3+\cdots +{\lambda }_{n-1}{\lambda }_n=a_{n-2}\\ \dots \\ {\lambda }_1{\lambda }_2\dots {\lambda }_n={(-1)}^n{a}_0={(-1)}^n\det 𝑨\end{array}}$

Fundamental set of solutions for hLSDE

• Solve the eigenvalue problem $𝑨𝒙=\lambda 𝒙$

• Form the set of solutions $𝒮=\{e^{{\lambda }_{1}t}{𝒙}_1,\dots ,e^{{\lambda }_{n}t}{𝒙}_n\}$

• Determine is the set of solutions is fundamental:

  • Does $c_1{e}^{{\lambda }_{1}t}{𝒙}_1+\dots +c_n{e}^{{\lambda }_{n}t}{𝒙}_n=𝟎$ imply $c_1=\cdots =c_n=0$? If so, $𝒮$ is a fundamental set

  • Compute the Wronskian

    ${\displaystyle W\left(t\right)=\left|\begin{array}{llll}{e}^{{\lambda }_{1}t}{x}_{11}& e^{{\lambda }_{2}t}{x}_{12}& \dots & e^{{\lambda }_{n}t}{x}_{1n}\\ e^{{\lambda }_{1}t}{x}_{21}& e^{{\lambda }_{2}t}{x}_{22}& \dots & e^{{\lambda }_{n}t}{x}_{2n}\\ ⋮& ⋮& \ddots & ⋮\\ e^{{\lambda }_{1}t}{x}_{n1}& e^{{\lambda }_{2}t}{x}_{n2}& \dots & e^{{\lambda }_{n}t}{x}_{\mathrm{nn}}\end{array}\right|=e^{{\lambda }_{1}t}\dots e^{{\lambda }_{n}t}\left|\begin{array}{llll}{x}_{11}& x_{12}& \dots & x_{1n}\\ x_{21}& x_{22}& \dots & x_{2n}\\ ⋮& ⋮& \ddots & ⋮\\ x_{n1}& x_{n2}& \dots & x_{\mathrm{nn}}\end{array}\right|}$

    and if $W\left(t\right)\ne 0$ for any $t$, $𝒮$ is a fundamental set of solutions

• Eigenvector matrix

  ${\displaystyle 𝑿=\left(\begin{array}{llll}{𝒙}_1& {𝒙}_2& \dots & {𝒙}_n\end{array}\right)=\left(\begin{array}{llll}{x}_{11}& x_{12}& \dots & x_{1n}\\ x_{21}& x_{22}& \dots & x_{2n}\\ ⋮& ⋮& \ddots & ⋮\\ x_{n1}& x_{n2}& \dots & x_{\mathrm{nn}}\end{array}\right)}$

Negative real eigenvalues

• Suppose ${\lambda }_1,\dots ,{\lambda }_n<0$, and $𝒮=\{e^{{\lambda }_{1}t}{𝒙}_1,\dots ,e^{{\lambda }_{n}t}{𝒙}_n\}$ is a fundamental set

  ${\displaystyle 𝒚=c_1{e}^{{\lambda }_{1}t}{𝒙}_1+\dots +c_n{e}^{{\lambda }_{n}t}{𝒙}_n}$

• Then ${lim}_{t\to \infty }𝒚\left(t\right)=𝟎$

(%i14) 

A: matrix([-7,4],[-5,2])$
p: factor(charpoly(A,lambda));

(%i3) 

y: matrix([y1(t)],[y2(t)])$ rhs: A.y$
dy: matrix(['diff(y1(t),t)],['diff(y2(t),t)])$
eq1: dy[1][1] = rhs[1][1]$
eq2: dy[2][1] = rhs[2][1]$
sys: [eq1,eq2]$
gsoln: desolve(sys,[y1(t),y2(t)])$
psoln: subst([y1(0)=1,y2(0)=0],gsoln);

(%i11) 

plot2d([parametric,rhs(psoln[1]),rhs(psoln[2]),
       [t,0,10]],[xlabel,"y1"],[ylabel,"y2"])$

(%i12) 

Positive real eigenvalues

• Suppose one eigenvalue is real positive, and $𝒮=\{e^{{\lambda }_{1}t}{𝒙}_1,\dots ,e^{{\lambda }_{n}t}{𝒙}_n\}$ is a fundamental set

  ${\displaystyle 𝒚=c_1{e}^{{\lambda }_{1}t}{𝒙}_1+\dots +c_n{e}^{{\lambda }_{n}t}{𝒙}_n}$

• Then generally $𝒚\left(t\right)\to \infty$ (see Lesson 20 for discussion)

(%i111) 

A: matrix([2,4],[4,2])$
p: factor(charpoly(A,lambda));

(%i113) 

y: matrix([y1(t)],[y2(t)])$ rhs: A.y$
dy: matrix(['diff(y1(t),t)],['diff(y2(t),t)])$
eq1: dy[1][1] = rhs[1][1]$
eq2: dy[2][1] = rhs[2][1]$
sys: [eq1,eq2]$
gsoln: desolve(sys,[y1(t),y2(t)])$
psoln: subst([y1(0)=1,y2(0)=0],gsoln);

(%i122) 

plot2d([parametric,rhs(psoln[1]),rhs(psoln[2]),
       [t,0,2]],[xlabel,"y1"],[ylabel,"y2"])$

Purely imaginary eigenvalues

• Suppose all eigenvalues are purely imaginary, $𝒮=\{e^{{\lambda }_{1}t}{𝒙}_1,\dots ,e^{{\lambda }_{n}t}{𝒙}_n\}$ is a fundamental set

  ${\displaystyle 𝒚=c_1{e}^{{\lambda }_{1}t}{𝒙}_1+\dots +c_n{e}^{{\lambda }_{n}t}{𝒙}_n}$

• Then $𝒚\left(t\right)$ exhibits cycles

(%i12) 

A: matrix([0,1],[-1,0])$
p: factor(charpoly(A,lambda));

(%i14) 

y: matrix([y1(t)],[y2(t)])$ rhs: A.y$
dy: matrix(['diff(y1(t),t)],['diff(y2(t),t)])$
eq1: dy[1][1] = rhs[1][1]$
eq2: dy[2][1] = rhs[2][1]$
sys: [eq1,eq2]$
gsoln: desolve(sys,[y1(t),y2(t)])$
psoln: subst([y1(0)=1,y2(0)=0],gsoln);

(%i22) 

plot2d([parametric,rhs(psoln[1]),rhs(psoln[2]),
       [t,0,6.3]],[xlabel,"y1"],[ylabel,"y2"])$

(%i23) 

Purely imaginary eigenvalues

• Suppose all eigenvalues are complex, $𝒮=\{e^{{\lambda }_{1}t}{𝒙}_1,\dots ,e^{{\lambda }_{n}t}{𝒙}_n\}$ is a fundamental set

  ${\displaystyle 𝒚=c_1{e}^{{\lambda }_{1}t}{𝒙}_1+\dots +c_n{e}^{{\lambda }_{n}t}{𝒙}_n}$

• Then $𝒚\left(t\right)$ exhibits spirals, inward towards zero for negative real part, outwards towards infinity for positive real part

(%i60) 

A: matrix([-1,-10],[10,-1])$
p: factor(charpoly(A,lambda));

(%i62) 

y: matrix([y1(t)],[y2(t)])$ rhs: A.y$
dy: matrix(['diff(y1(t),t)],['diff(y2(t),t)])$
eq1: dy[1][1] = rhs[1][1]$
eq2: dy[2][1] = rhs[2][1]$
sys: [eq1,eq2]$
gsoln: desolve(sys,[y1(t),y2(t)])$
psoln: subst([y1(0)=1,y2(0)=0],gsoln);

(%i70) 

plot2d([parametric,rhs(psoln[1]),rhs(psoln[2]),
       [t,0,24]],[xlabel,"y1"],[ylabel,"y2"])$

(%i71)